# Using definition of Cartier divisors on $\mathbb{P}^1$.

Suppose I use the following definition of a Cartier divisor: a global section of the quotient sheaf $$\mathcal{K}^*/\mathcal{O}^*$$.Because of the property of quotient sheaves a Cartier divisor on $$X$$ can be described by giving an open cover $$\{U_i\}$$ of $$X$$ and $$f_i \in \Gamma(U_i, \mathcal{K}^*)$$ such that $$f_i / f_j \in \Gamma(U_i \cap U_j, \mathcal{O}^*)$$

I thought it would be interesting to use this definition to figure out the divisors on $$\mathbb{P}^1$$. In pariticular, I want to show that every divisor is a multiple of the hypersurface $$x_0=0$$. This would reprove that $$D \sim dH$$.

I am getting stuck trying to figure out what the global sections of this sheaf are. However, even if I were to find the global sections I am not exactly sure how I would relate them to a hypersurface.

Would I just compute the vanishing set of the global section I find?

Its still a bit confusing to me that Cartier divisors are defined as global sections of a quotient sheaf and Weil divisors are closed subschemes of codimension $$1$$ of our scheme.

Let's look at the global section of this sheaf. Element of $\mathscr K^*$ are exactly non-identically zero rational functions $\frac{f}{g}$, and $\mathcal O^*$ is simply the global section of non-identically zero holomorphic functions, i.e $\mathcal O^*(\Bbb P^1) =k^*$.
Assuming $k$ is algebraically closed, any such $f/h$ will be product of some linear factors quotiented by product of some other linear factors, i.e $f/h$ is equivalent to the data of the zeroes and the poles with multiplicities, which is exactly the data of a divisor $D = \sum_{p \in \Bbb P^1} D(p) p$. For example, the Cartier divisor $$[\frac{(x_0 - x_1)(x_0 + x_1)}{x_1^2}]$$ will correspond to the Weil divisor $D = [1:1] + [1:-1] - 2 \cdot [1:0]$.
Now, it's easy to check that if $D,D'$ have the same degree then there is a rational function $f$ with $\text{div}(f) = D - D'$. Taking $D' = \deg(D) H$ shows that any divisor is equivalent to some multiple of $H$.